Answer
0.44
Work Step by Step
Use equation 18–3 to find the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
$$\frac{R_{Al}}{R_{Cu}}=\frac{\rho_{Al}\mathcal{l}_{Al}d_{Cu}^2}{\rho_{Cu}\mathcal{l}_{Cu}d_{Al}^2}$$
$$=\frac{(2.65\times10^{-8}\Omega \cdot m)(10.0m)(0.0018m)^2}{(1.68\times10^{-8}\Omega \cdot m)(24.0m)(0.0022m)^2}=0.44$$
